Answer: 9.9 grams
Step-by-step explanation:
To calculate the moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}](https://img.qammunity.org/2020/formulas/chemistry/middle-school/7f4c6armgb5j4osly6yzny62ubzrugwqj8.png)
a) moles of
![H_2](https://img.qammunity.org/2020/formulas/chemistry/high-school/qm59xqb9s1bnftac71jms9xnzoqfea6xdm.png)
![\text{Number of moles}=(8.150g)/(2g/mol)=4.08moles](https://img.qammunity.org/2020/formulas/chemistry/high-school/3novoz7d3m5lbleln5yq9kec0oeugxu8zg.png)
b) moles of
![C_2H_4](https://img.qammunity.org/2020/formulas/chemistry/high-school/tva0g3pdfnvh6hmvgm1iz6hlrbi0b6n12x.png)
![\text{Number of moles}=(9.330g)/(28g/mol)=0.33moles](https://img.qammunity.org/2020/formulas/chemistry/high-school/bzht68wk3ovnolb6crc8iwxbsu7ac5p1zn.png)
![H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)](https://img.qammunity.org/2020/formulas/chemistry/high-school/e2r2mzyzxrmts2iyi6nrvpyntktteveto2.png)
According to stoichiometry :
1 mole of
combine with 1 mole of
Thus 0.33 mole of
will combine with =
mole of
Thus
is the limiting reagent as it limits the formation of product.
As 1 mole of
give = 1 mole of
![C_2H_6](https://img.qammunity.org/2020/formulas/chemistry/middle-school/u6uhywmnr7sb6yj60rfkjtte8y6w1dxt8b.png)
Thus 0.33 moles of
give =
of
![C_2H_6](https://img.qammunity.org/2020/formulas/chemistry/middle-school/u6uhywmnr7sb6yj60rfkjtte8y6w1dxt8b.png)
Mass of
![C_2H_6=moles* {\text {Molar mass}}=0.33moles* 30g/mol=9.9g](https://img.qammunity.org/2020/formulas/chemistry/high-school/pqa5b0435spfe6t9yg761xoqw7lkq261gm.png)
Thus theoretical yield (g) of
produced by the reaction is 9.9 grams